Show that the 9-point Circle of a triangle satisfies the cyclic ordered product of segments
$$ \dfrac{BP}{PC}\times \dfrac{CQ} {QA}\times \dfrac{AR}{RB}=1 $$
Show that it could be included in a generalization Menelaus's theorem as a special case of a proposition where the product of ratios is any real number.
Context was yesterday's attempt to find such triple ratio product locus with my following logic for this particular case.
The circle has equal segments on its sides, satisfying the generalization
At foot of perpendicular through Ortho-center the opposite sides are partitioned as:
$$ \dfrac{a \cos B}{b \cos A}\cdot\dfrac{b \cos C}{c \cos B}\cdot \dfrac{c \cos A}{a \cos C} =1$$
which also satisfies the generalization.
This has nothing to do with the nine-point circle. The three points in the sides are the feet of the altitudes. The altitidues are concurrent and, as such, your triple product is a special case of Ceva's theorem.